Solve (x+1)(x+1)^2-3(x+1)+3]-1

$$( x + 1 ) ( x + 1 ) ^ { 2 } - 3 ( x + 1 ) + 3 ] - 1$$

Evaluate

$$\left(x+1\right)^{3}-3x$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$\left(x+1\right)^{3}-3\left(x+1\right)+3$$

Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(x+1\right)^{3}.

$$x^{3}+3x^{2}+3x+1-3\left(x+1\right)+3$$

Use the distributive property to multiply -3 by x+1.

$$x^{3}+3x^{2}+3x+1-3x-3+3$$

Combine 3x and -3x to get 0.

$$x^{3}+3x^{2}+1-3+3$$

Subtract 3 from 1 to get -2.

$$x^{3}+3x^{2}-2+3$$

Add -2 and 3 to get 1.

$$x^{3}+3x^{2}+1$$

Expand

$$x^{3}+3x^{2}+1$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$\left(x+1\right)^{3}-3\left(x+1\right)+3$$

Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(x+1\right)^{3}.

$$x^{3}+3x^{2}+3x+1-3\left(x+1\right)+3$$

Use the distributive property to multiply -3 by x+1.

$$x^{3}+3x^{2}+3x+1-3x-3+3$$

Combine 3x and -3x to get 0.

$$x^{3}+3x^{2}+1-3+3$$

Subtract 3 from 1 to get -2.

$$x^{3}+3x^{2}-2+3$$

Add -2 and 3 to get 1.

$$x^{3}+3x^{2}+1$$